Current and circuits

Reading: Chapters 25, 26, CURRENT and DC CIRCUITS

Study guide: Chapters 26 & 27

  1. Current - microscopic picture
  2. Ohm's law
  3. Resistance and resistivity
  4. Electromotive Force (EMF)
  5. Power
  6. Solving circuits
  7. Time-dependent circuits (w capacitors)

Current

Charges move on average with a constant drift velocity in a wire in response to an applied electric field. ...Not an acceleration??

In a metal, electrons are the mobile charges. They are constantly "bumping into" the positive ions of the host material, which are fixed because of their bonds to neighboring ions in a crystal lattice.

...is the rate at which charge passes through an imaginary cross-section per unit time: $$I=\frac{\Delta Q}{\Delta t}$$ Units: $\frac{C}{s}\equiv A$: "Amps".

Direction of current is the direction that positive charges would move in response to $\myv E$--opposite of direction of motion of negative charges.

$$\Delta Q = ( n )(v_D\Delta t\,A)(q)$$

Resistance

Many materials have linear relation between current $I$ and potential difference $V$:$$I=\frac{1}{R} V.$$

More variations through algebra:$$R=\frac{V}{I};\ \ V=IR.$$

This is Ohm's law. $R$ has units of $V/A\equiv \Omega$, "Ohms".

Wires have low resistance. [Much current or little current at the same V?]

Resistance depends on the geometry of wire, and the material it's made of: $$R=\rho \frac{l}{A}$$ where this time $\rho$ is the resistivity of the material (units $\Omega m$).

See this reminder/animation of resistance and resistivity.

[Example: Household wiring]

Different kinds of conductors

Metals - increasing resistance (same number of charge carriers) with increasing temperature:$$\rho(T)=\rho_0\left[ 1+\alpha(T-T_0)\right].$$

Semiconductors - More charge carriers with increasing temperature (pothole model), resistance drops with increasing temperature.

Superconductors - Generally metallic at high temps, but below a critical temperature, the resistance drops to 0.

Dielectrics (insulators) - Resistance is effectively infinite up until some breakdown field.

Electromotive Force - EMF

Resistance causes charge carriers to lose energy, so they will stop moving unless something gives them more electrical potential energy

A device that transforms energy from some other form into electrical potential energy is called a source of electromotive force. (aka "battery").

Most batteries give each charge that passes through the same increase in energy. The work per unit charge that such a device does on charges is designated $\cal{E}$ and has units of Volts.

[Graph of the life of an electron.]

Real batteries

 

Lead-acid battery

3 cells in series to get 12 V.

So, we model batteries as having some internal resistance, $r$, and so the voltage across the terminals is $$V_a - V_b={\cal E}-Ir$$

[What is the voltage across the terminals when no current is flowing?]

[Figure out $I$ in the accompanying circuit--related to SG problem 26.5]

Power

The change in energy of a particle of charge $q$ which is 'falling' through a potential difference of $V$ in a resistor is $\Delta U=-qV$.

Power is energy change per unit time: $$P=\frac{\Delta U}{\Delta t}=\frac{q}{\Delta t} V = I\cdot V$$

As long as we are dealing with ohmic devices, we can use $V=IR$ to write the power in several different ways...

$$P=I^2R=V^2/R.$$

This energy is lost to the circuit, to heating of the positive ions: "Joule heating".

For a battery (source of EMF), the change of energy is $q{\cal E}$ and the power that it supplies to the circuit is$$P={\cal E}I.$$

[Resistance of a 1000 W hair dryer?]

Solving circuits

Some observations you have perhaps made in lab

Now figure out...

Resistors in series and parallel

Series:

$$\begineq V&=&V_1+V_2+V_3+...\\ &=&IR_1+IR_2+IR_3+...\\ &=&I(R_1+R_2+R_3) = I R_\text{eff}\endeq$$

So...$$R_\text{eff}= R_1+R_2+R_3+...$$

Parallel:

and then$$\begineq I&=&I_1+I_2+I_3+...\\ &=& V/R_1+V/R_2+V/R_3+...\\ &=& V\left( \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...\right)\\ &=&V/R_\text{eff}\endeq$$

So:$$\frac{1}{R_\text{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...$$

[Resistance of three 10$\Omega$ resistors in parallel?]

[Resistance of cireuit ]

Ideal meters

Voltmeter--no current should flow...

Ammeter--no voltage drop as current flows through...

Charging/discharging a capacitor

Charge up a capacitor with a battery of voltage ${\cal E}$. The capacitor starts out with:

Now, disconnect the battery, and connect a resistor $R$ across the poles of the capacitor. Initially, the current that flows through the resistor is:$$i_0=\frac{v_0}{R}$$ But then, what happens next? [Graph $i(t)$ qualitatively. Graph charge on the capacitor $q(t)$ qualitatively]

Convention: time-dependent quantitities are written in lower case. Current in the circuit, voltage and charge on the capacitor are written $i(t)$, $v(t)$, $q(t)$.

Getting more precise: $$i=-\frac{\Delta q}{\Delta t}\to -\frac{dq}{dt}.$$ At all times, the resistor obeys Ohm's law, so...

$$\frac{dq}{dt}=-i(t) = -\frac{v(t)}{R}.$$

The charge that's flowing in the circuit is charge which is being lost from the capacitor. Remembering $Q=CV$, we can write the voltage in terms of the charge on the capacitor as...

$$\frac{dq}{dt}=-i(t)=-\frac{v(t)}{R}=-\frac{q(t)}{RC}.$$

Oh, this is a differential equation for $q(t)$: $$\frac{dq}{dt}=-\frac{q}{RC}$$ Based on our graph of $q(t)$ let's guess that the solution is a decaying exponential: $$q(t)=q_0 e^{-kt}$$

Try this out in our differential equation: $$\frac{dq}{dt}= q_0(-k)e^{-kt}$$

So we have a solution as long as $k=1/(RC)$...$$q(t)=q_0 e^{-t/(RC)}$$

$$\Rightarrow v(t)=\frac{q(t)}{C}=\frac{q_0}{C}e^{-t/RC}=v_0e^{-t/RC}.$$

$$\Rightarrow i(t)=\frac{v(t)}{R} = \frac{v_0}{R}e^{-t/RC}=i_0e^{-t/RC}.$$

Charging *up* an uncharged ($v_0=0$) capacitor, with a resistance in the circuit, it turns out that these quantities approach their final values exponentially as...

$$v(t)={\cal E}\left(1-e^{-t/RC}\right)$$ $$q(t)=Cv(t)={\cal E}C\left(1-e^{-t/RC}\right)$$ $$i(t)=-\frac{\cal E}{R}e^{-t/RC}$$